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evablogger [386]
3 years ago
12

A roller coaster car has a mass of 400kg and a speed of 15m/s What will be the P.E of this roller coaster car at its highest poi

nt ,where KE =0 at that point
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

3000J

Explanation:

Given parameters:

mass  = 400kg

speed  = 15m/s

Unknown:

P.E at the highest point of the roller coaster = ?

Solution:

Due to the law of conservation of energy, the potential energy at the highest point can be solved using the formula of the kinetic energy

  Potential energy  = \frac{1}{2} mv²

m is the mass

v is the velocity

    Potential energy =  \frac{1}{2}  x 400 x 15  = 3000J

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posledela

Red clothes look red  because they REFLECT the red light, and absorb light of other colors.

7 0
3 years ago
Derive the value of electric field due to a uniform sphere of charge.
Deffense [45]
<h2>Hey there!</h2>

The Force "F" applied on the unit electric charge "q" at a point describes the electric field.

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4 0
2 years ago
Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl
lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

 Weight of body = 15 N downwards = 15 N

 Buoyancy force on body = 17 N upwards = -17 N

 Total force on body = 15 - 17 = -2 N = 2 N upwards

 So, the body will float.

Object will float.

8 0
3 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
Check out this app! It's millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02Ma
brilliants [131]

Answer:

ok ty

Explanation:

5 0
3 years ago
Read 2 more answers
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